Ch 6.5: Inverse trigonometric functions In this section, we will 1. define an one-to-one function 2. look at the conditions on which inverse functions exist 3. define and graph inverse functions for the six (6) trig. functions 4. look at common examples
One-to-One function A function f is one-to-one if every element in the range corresponds to only one element in the domain. That is, Definition (One-to-One function) A function f is one-to-one if x 1 x 2 f (x 1 ) f (x 2 ) Question) Are these functions 1-1?
Is the following function 1-1? Here s the graph of y = 1.
The Horizontal line test Recall that a function is one-to-one if for all x 1 and x 2, x 1 x 2 f (x 1 ) f (x 2 ). Thus, a function is not one-to-one if there are x 1 and x 2 with x 1 x 2 f (x 1 ) = f (x 2 ). That is, if there is a horizontal line that intersects the graph of the function two or more points, the function fails the horizontal line test and is not 1-1. Q: Which function(s) pass(es)/fail(s) the horizontal line test?
Why one-to-one? Given y = f (x), its inverse function, f 1, is when we solve for x and then interchange x and y. Thus, if a function f is 1-1 (i.e., if it passes the horizontal line test), the graph with x and y switched will pass the vertical line test, which will mean that f 1 exists as a function. (Note: All functions must pass the vertical line test.) Does the function pass the horizontal test? Does the inverse function exist?
Inverse of f (x) = x 2? Does the function pass the horizontal test? Does the inverse function exist? (If not, how can we make it exist?)
Graphs of inverse functions To find the graph of the inverse function, draw the (invisible) y = x line and flip everything about that line.
Graphs of inverse functions What about y = x 2 on [0, )?
A note about inverse functions Since we interchange x and y to find the inverse function of f the domain of f 1, (D(f 1 )), is the range of f, or (R(f )). the range of f 1, (R(f 1 )), is the domain of f, or (D(f )). Example) Given f = x 2 on [0, 2], find the domain and range of its inverse.
Inverse trig functions and their applications Question: Which of the sine, cosine and tangent functions pass(es) the horizontal test? That is, which of the three functions are 1-1?
Let s restrict the domain of the trig functions. Here s the sine function: How can we make it 1-1?
How about cosine? Here s the cosine function: How can we make it 1-1?
How about tangent? Here s the tangent function: How can we make it 1-1?
Summary- How to make these functions 1-1? For f (x) = sin x, restrict its domain to [ π 2, π 2 ]. For f (x) = cos x, restrict its domain to [0, π]. For f (x) = tan x, restrict its domain to ( π 2, π 2 ).
y = arcsin x = sin 1 (x) Here s the sine function restricted on [ π 2, π 2 ]. If we interchange x and y or flip it about the y = x, then
y = arcsin x = sin 1 (x) Domain: [ 1, 1] (= range of sin x) Range: [ π 2, π ] 2 ( = domain of sin x) sin(sin 1 x) = x for x sin 1 (sin x) = x for x
y = arccos x = cos 1 (x) Here s the cosine function restricted on [0, π]. If we interchange x and y or flip it about the y = x, we have
y = arccos x = cos 1 (x) Domain: [ 1, 1] (= range of cos x) Range: [0, π] ( = domain of cos x) cos(cos 1 x) = x for x cos 1 (cos x) = x for x
y = arctan x = tan 1 (x) Domain: (= range of tan x) Range: ( = domain of tan x) tan(tan 1 x) = x for tan 1 (tan x) = x for
Example 1 Solve each equation ( for y without the use of a calculator. 1. y = sin 1 ) 3 2 2. y = arcsin ( 1 ) 2 3. y = sin 1 (2)
Example 2 Evaluate each expression: 1. y = sin [ sin 1 ( )] 1 2 2. y = arcsin [ sin ( )] 3π 4 3. y = sin 1 [ sin ( )] 5π 6
Example 3 Evaluate the inverse cosine function for the values given: 1. y = cos 1 (0) ( 2. y = arccos 3. y = cos 1 (π) 3 2 )
Example 4 Evaluate each expression: 1. y = cos [ cos 1 ( 0.2) ] 2. y = arccos [ cos ( )] π 12 3. y = cos 1 [ cos ( )] 4π 3
Example 5 Evaluate each expression: 1. y = tan 1 ( 3) 2. y = arctan [tan (.89)] 3. y = tan 1 [ tan ( )] 7π 6
Example 6 Simplify each expression: 1. y = tan [ sin 1 ( 1 )] 2 2. y = sin 1 [ cos ( )] π 3
Example 7 Evaluate the expression tan [ sin 1 ( 4 5)].
Example 8 Evaluate the following expression as an algebraic function of x: sin [ cos 1 2x ]. Assume 1 2 x 1 2.
Example 9 ( Evaluate the expression tan [cos 1 x x 2 +16 )]. Assume x > 0.
Inverse functions for secant (Not covered in textbook) First, here s the graph of the secant function. then, we may restrict our domain to [0, π 2 ) ( π 2, π], so that y = sec 1 x can be defined.
Graph of y = sec 1 x Insert graph of the inverse function here. Domain Range
Inverse functions for cosecant First, here s the graph of the cosecant function. then, we may restrict our domain to [ π 2, 0) (0, π 2 ], so that y = csc 1 x can be defined.
graph of y = csc 1 x Domain Range
Inverse functions for cotangent First, here s the graph of the cotangent function. then, we may restrict our domain to either [ π 2, 0) (0, π 2 ) or (0, π), so that the graph of the y = cot 1 x is then
y = cot 1 x or Domain Range
Example 10 Evaluate the following: 1. csc 1 (2) ( ) 2. sec 1 3 2 ( ) 3. cot 1 3 1